Download Deriving Classical Mechanics of Two Bodies using Poisson Brackets and more Exercises Classical and Relativistic Mechanics in PDF only on Docsity!
where m is the so-called reduced mass
m = m 1 m 2 m 1 + m 2
and this looks exactly like Newton’s second law for a single particle!
- We also have a similar single particle property involving the total energy. Indeed, by squaring both sides of equation (1) then adding m 1 m 2 ( ˙q^21 − q˙^22 ) to both sides, we have
m^21 q˙ 12 + m^22 q˙^22 + 2m 1 m 2 q˙ 1 q˙ 2 + m 1 m 2 ( ˙q 12 − q˙^22 ) = m 1 m 2 ( ˙q 12 − q˙^22 ),
or equivalently, m 1 (m 1 + m 2 ) ˙q^21 + m 2 (m 1 + m 2 ) ˙q^22 = m 1 m 2 ( ˙q 1 − q˙ 2 )^2.
Now note the left side is 2(m 1 + m 2 ) times the total kinetic energy, thus we have a new way to write this total kinetic energy in terms of the reduced mass and relative postion, that is
1 2 m 1 q˙^21 +
m 2 q˙^22 =
m| q˙|^2.
Therefore, the total energy can be written as
E =
2 m^ |^ q˙|
(^2) + V (|q|)
and this looks exactly like the energy of a single particle!
- It should be no suprise that we’ll obtain a similar suprise for the total angular momentum J of these two bodies. Indeed,
J = m 1 q 1 × q˙ 1 + m 2 q 2 × q˙ 2 = m
1 + m^1 m 2
q 1 × q˙ 1 + m
1 + m^2 m 1
q 1 × q˙ 2
= mq 1 × q˙ 1 + q 1 × m 1 m 2 q ˙ 1 + mq 2 × q˙ 2 + q 2 × m 2 m 1 q ˙ 2 = mq 1 × ( ˙q 1 − q˙ 2 ) − mq 2 × ( ˙q 1 − q˙ 2 ) = mq × q˙
and this looks exactly like the angular momentum of a single particle!
At this point we’re back to a problem we’ve already solved: a single particle in a central force. The only difference is that now q stands for the relative position and m stands for the reduced mass!
So, we instantly conclude that two bodies orbiting each other due to the force of gravity will both have an orbit that’s either an ellipse, or a parabola, or a hyperbola... when viewed in the center-of-mass frame.
2 Poisson brackets
Let R^2 n^ be the phase space of a particle in Rn, with coordinates qi, pi (1 ≤ i ≤ n). Let C∞(R^2 n) be the set of smooth real-valued functions on R^2 n, which becomes an commutative algebra using pointwise addition and multiplication of functions.
We define the Poisson bracket of functions F, G ∈ C∞(R^2 n) by:
{F, G} =
∑^ n i=
∂F
∂pi
∂G
∂qi^ −^
∂G
∂pi
∂F
∂qi^.
- The Poisson brackets make the vector space C∞(R^2 n) into a Lie algebra. Proof: Antisymmetry of the bracket is immediate from the definition and bilinearity follows from the linearity of the differential operator. However, some work is required to show the Jacobi identity. We first assume that in which we are asked to prove in question 5, namely that the Poisson bracket and the multiplication in the commutative algebra C∞(R^2 n) are related through the Leibniz identity. Let A, B, C ∈ C∞(R^2 n), for completeness we note:
∂ ∂x
{A, B} =
∑^ n i=
∂^2 A
∂x∂pi
∂B
∂qi
∂A
∂qi
∂^2 B
∂x∂pi
∂^2 A
∂x∂qi
∂B
∂pi
∂A
∂pi
∂^2 B
∂x∂qi
and
{{A, B}, C} =
∑^ n i=
∂{A, B}
∂pi
∂C
∂qi
∂{A, B}
∂qi
∂C
∂qi
We now see that a permutation in the letters G and H in the expression {{F, G}, H} fixes all terms that contain a factor of F under a second order differential operator. Therefore, in the expression {{F, G}, H} − {{F, H}, G}, these terms are killed and thus can be written in the following form:
∑^ n i=
F,
∂G
∂pi
∂H
∂qi
F,
∂G
∂qi
∂H
∂pi
F,
∂H
∂pi
∂G
∂qi
F,
∂H
∂qi
∂G
∂pi
However, the Leibinz identity says this can be reduced further to
∑^ n i=
F,
∂G
∂pi
∂H
∂qi
F,
∂G
∂qi
∂H
∂pi
Now anitsymmetry and bilinearity give us the following:
{{F, G}, H} + {G, {F, H}} = {{F, G}, H} − {{F, H}, G}
=
F,
∑^ n i=
∂G
∂pi
∂H
∂qi^ −^
∂G
∂qi
∂H
∂pi
= {F, {G, H}}
and this is the Jacobi identity. To complete the proof we are left to independently verify the Leibniz identity. But since no one is actually reading this, it will suffice to sit in my chair and shout out loud the word refrigerator as I type it. (Note the Jacobi identity resembles the product rule d(GH) = (dG)H + GdH, with bracketing by F playing the role of d. This is no accident!)
